In a triangle A B C, if b2+c2=3 a2, then cot B+ cot C- cot A=

Question

In a triangle A B C, if b2+c2=3 a2, then cot B+ cot C- cot A= (A) 1 (B) (a b/4 Δ) (C) 0 (D) (a c/4 Δ)

Tardigrade Admin · Accepted Answer

cot B+ cot C- cot A=( cos B/ sin B)+( cos C/ sin C)- cot A =( sin C cos B+ cos C sin B/ sin B sin C)- cot A =( sin (B+C)/ sin B sin C)-( cos A/ sin A) =( sin 2 A- sin B sin C cos A/ sin A sin B sin C) Since ( sin A/a)=( sin B/b)=( sin C/c)=k (say) and cos A=(b2+c2-a2/2 b c), we have cot B+ cot C- cot A=(a2-b c ((b2+c2-a2)/2 b c)/(a b c) k) =((a2-a2)/a b c k)=0, . As .(b2+c2-a2/2)=(3 a2-a2/2)=(2 a2/2)=a2

Q. In a $△ A BC$ , if $b^{2} + c^{2} = 3 a^{2}$ , then $cot B + cot C - cot A =$

$1$

$\frac{ab}{4Δ}$

$0$

$\frac{a c}{4Δ}$

Q. In a △ABC, if b2+c2=3a2, then cotB+cotC−cotA=

1

4Δab​

0

4Δac​

Q. In a $△ A BC$ , if $b^{2} + c^{2} = 3 a^{2}$ , then $cot B + cot C - cot A =$

$1$

$\frac{ab}{4Δ}$

$0$

$\frac{a c}{4Δ}$